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anonymous

  • 5 years ago

please help! find the absolute max and min of f(x)=x^3-12x on [0,4]

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  1. anonymous
    • 5 years ago
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    answer : 2

  2. anonymous
    • 5 years ago
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    y(dx/dy) = 3x^2 -12 getting the critical numbers: 3x^2 - 12 = 0 3x^2 = 12 x^2 = 12/3 x^2 = 4 x = +-2 substitute it to the original value y1=(2)^3-12(2) y1=-16 y2=(-2)^3-12(-2) y2=16 get the y value of the interval y3 = (0)^3 - 12(0) y3 =0 y4 = (4)^3 - 12(4) y4 = 16 biggest is 16 now we know that abs max is 16 at x = 4 and x = -2 smalles is -16 so abs min is -16 at x=2

  3. anonymous
    • 5 years ago
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    find the derivative of f(x) and solve it when it's equal to 0 f'(x)=3x^2-12 \[3x^2-12=0 \implies 3x^2=12 \implies x^2=4\] x=2 or x=-2 but -2 is not in [0,4] all you have to do now is to check the values of the function at x=2, and at the border points x=0, x=4 the maximum among them is the maximum in the interval, and same is applied to the minimum

  4. anonymous
    • 5 years ago
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    oh yea i forgot x= -2 cant be considered~~ but nevertheless 16 is still the abs maximum ^^

  5. anonymous
    • 5 years ago
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    f(2)=2^3-12(2)=8-24=-16 f(0)=0 f(4)=4^3-12(4)=16 so the absolute maximum is 16 at x=4 and the absolute minimum is -16 at x=2

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